$\overline{AB} = 26$ $\overline{AC} = {?}$ $A$ $C$ $B$ $26$ $?$ $ \sin( \angle BAC ) = \dfrac{12}{13}, \cos( \angle BAC ) = \dfrac{5}{13}, \tan( \angle BAC ) = \dfrac{12}{5}$
$\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{26} $ Since we have already been given $\cos( \angle BAC )$ , we can set up a proportion to find $\overline{AC}$ $ \cos( \angle BAC ) = \dfrac{5}{13} = \frac{\overline{AC}}{26}$ Simplify. $\overline{AC} = 10$